POWER SYSTEM STABILITY
POWER SYSTEM STABILITYLESSON SUMMARY-1:1. Introduction2. Classification of Power System Stability3. Dynamic Equation of Synchronous MachinePower system stability involves the study of the dynamics of the power systemunder disturbances. Power system stability implies that its ability to return tonormal or stable operation after having been subjected to some form ofdisturbances.From the classical point of view power system instability can be seen as loss ofsynchronism (i.e., some synchronous machines going out of step) when the systemis subjected to a particular disturbance. Three type of stability are of concern:Steady state, transient and dynamic stability.Steady-state Stability:Steady-state stability relates to the response of synchronous machine to agradually increasing load. It is basically concerned with the determination of theupper limit of machine loading without losing synchronism, provided the loadingis increased gradually.Dynamic Stability:Dynamic stability involves the response to small disturbances that occur on thesystem, producing oscillations. The system is said to be dynamically stable iftheses oscillations do not acquire more than certain amplitude and die out quickly.If these oscillations continuously grow in amplitude, the system is dynamicallyunstable. The source of this type of instability is usually an interconnectionbetween control systems.Transient Stability:Transient stability involves the response to large disturbances, which may causerather large changes in rotor speeds, power angles and power transfers. Transientstability is a fast phenomenon usually evident within a few second.
Power system stability mainly concerned with rotor stability analysis. For thisvarious assumptions needed such as: For stability analysis balanced three phase system and balanced disturbancesare considered. Deviations of machine frequencies from synchronous frequency are small. During short circuit in generator, dc offset and high frequency current arepresent. But for analysis of stability, theses are neglected. Network and impedance loads are at steady state. Hence voltages, currentsand powers can be computed from power flow equation.Dynamics of a Synchronous Machine :The kinetic energy of the rotor in synchronous machine is given as:KE 1/2Jws2 x10 -6 MJoule . . (1)WhereJ rotor moment of inertia in kg-m2ws synchronous speed in mechanical radian/sec.Speed in electrical radian iswse (P/2) ws rotor speed in electrical radian/sec . .(2)WhereP no. of machine polesFrom equation (1) and (2) we getKE orWhereKE M MJ . . (3)MJ moment of inertia inMJ.sec/elect. radian . (4)We shall define the inertia constant H, such thatGH KE WhereMJ . . . (5)G three-phase MVA rating (base) of machineH inertia constant in MJ/MVA or MW.sec/MVA
From equation (5), we can write,orM MJ.sec/elect. radian . (6)M MJ.sec/elect. degree . . (7)M is also called the inertia constant.Assuming G as base, the inertia constant in per unit isorM(pu) Sec2/elect.radian . (8)M(pu) Sec2/elect.degree . (9)LESSON SUMMARY-2:1.2.3.4.Swing equationMulti machine systemMachines swinging in unison or coherentlyExamplesSwing Equation:-PeTePmGENERATORTmWs(Fig.-1 Flow of power in a synchronous generator)Consider a synchronous generator developing an electromagnetic torque Te(anda corresponding electromagnetic power Pe) while operating at the synchronousspeed ws. If the input torque provided by the prime mover, at the generator shaft isTi, then under steady state conditions (i.e., without any disturbance).
Te Ti . (10)Here we have neglected any retarding torque due to rotational losses. Thereforewe haveTe ws Ti ws . (11)Te ws - Ti ws Pi - Pe 0 . (12)AndWhen a change in load or a fault occurs, then input power Pi is not equal to Pe.Therefore left side of equation is not zero and an accelerating torque comes intoplay. If Pa is the accelerating (or decelerating) power, then . (13)Pi- Pe WhereD damping coefficientθe electrical angular position of the rotorIt is more convenient to measure the angular position of the rotor with respectto a synchronously rotating frame of reference. Letδ θe -ws.t . (14) . (15)SoWhere δ is power angle of synchronous machine.Rotor FieldδWsReference rotatingaxisθeReference axis(Fig.2 Angular Position of rotor with respect to reference axis)Neglecting damping (i.e., D 0) and substituting equation (15) in equation (13)we get
MW . (16)Using equation (6) and (16), we getMW (17)Dividing throughout by G, the MVA rating of the machine,pu . (18) . (19)Wherepu . (20)orEquation (20) is called Swing Equation. It describes the rotor dynamics for asynchronous machine. Damping must be considered in dynamic stability study.Multi Machine System:In a multi machine system a common base must be selected. LetGmachine machine rating (base)Gsystem system baseEquation (20) can be written as: (21)Sopu on system base . (22) .(23)Where machine inertia constant in system baseMachines Swinging in Unison (Coherently) :Let us consider the swing equations of two machines on a common systembase, i.e., . . (24)
. . . (25)Since the machines rotor swing in unison,δ 1 δ 2 δ . .(26)Adding equations (24) and (25) and substituting equation (26), we get . (27)WherePi Pi1 Pi2Pe Pe1 Pe2Heq H1 H2Equivalent inertia Heq can be expressed as: . (28)Example1:A 60 Hz, 4 pole turbo-generator rated 100MVA, 13.8 KV has inertia constantof 10 MJ/MVA.(a) Find stored energy in the rotor at synchronous speed.(b) If the input to the generator is suddenly raised to 60 MW for an electricalload of 50 MW, find rotor acceleration.(c) If the rotor acceleration calculated in part (b) is maintained for 12 cycles,find the change in torque angle and rotor speed in rpm at the end of thisperiod.(d) Another generator 150 MVA, having inertia constant 4 MJ/MVA is put inparallel with above generator. Find the inertia constant for the equivalentgenerator on a base 50 MVA.Solution:(a) Stored energy GH 100MVA x 10MJ/MVA 1000MJ(b) Pa Pi-Pe 60-50 10MWWe know, M MJ.sec/elect.deg.\
Now 10 108 elect.deg./sec2So, α acceleration 108 elect.deg./sec2(c) 12 cycles 12/60 0.2sec. Change in δ ½ α.(Δt)2 ½.108.(0.2)2 2.16 elect.degNow α 108 elect.deg./sec2 60 x (108/360ᵒ) rpm/sec 18 rpm/secHence rotor speed at the end of 12 cycles () rpm 1803.6 rpm.(d) Heq 32MJ/MVALESSON SUMMARY-3:1. Power flow under steady state2. Steady-state Stability3. ExamplesPower Flow under Steady State:Consider a short transmission line with negligible resistance.VS per phase sending end voltageVR per phase receiving end voltageVs leads VR by an angle δx reactance of per transmission lineI jxVs-VR-(Fig.3-A short transmission line)
On the per phase basis power on sending end,SS PS j QS VSI* . . (29)From Fig.3 I is given asI or . . (30)I* From equation (29) and (30), we get . (31)SS NowVR VR VS VS so, VR VR* VR VS Equation (31) becomesSS PS jQS SoandδPS δδ . . (32) . . . (33)Qs Similarly, at the receiving end we haveSR PR j QR VRI* . (34)Proceeding as above we finally obtainQR PR . (35) . . (36)Therefore for lossless transmission line,PS P R . . (37)In a similar manner, the equation for steady-state power delivered by a losslesssynchronous machine is given by
Pe P d δ (38) Where Eg is the rms internal voltage, Vt is the rms terminal voltage, xd is thedirect axis reactance (or the synchronous reactance in a round rotor machine) and δis the electrical power angle.Steady-state Stability:The steady state stability limit of a particular circuit of a power system definedas the maximum power that can be transmitted to the receiving end without loss ofsynchronism.Now consider equation (18), (39)WhereAndPe δ . (40)Let the system be operating with steady power transfer ofwith torqueangle . Assume a small increment ΔP in the electric power with the input fromthe prime mover remaining fixed atcausing the torque angle to change to(. Linearizing the operating point (we can write . (41)The excursions ofare then described by . . (42) . (43)oror[Mp2 Wherep . . (44)
The system stability to small changes is determined from the characteristicequationMp2 Where two roots areAs long asp 0 . (45) . (46)is positive, the roots are purely imaginary and conjugateand system behavior is oscillatory about . Line resistance and damper windingsof machine cause the system oscillations to decay. The system is therefore stablefor a small increment in power so long asWhenδδis negative, the roots are real, one positive and the othernegative but of equal magnitude. The torque angle therefore increases withoutbound upon occurrence of a small power increment and the synchronism is soonlost. The system is therefore unstable forδis known as synchronizing coefficient. This is also called stiffness ofsynchronous machine. It is denoted as Sp. This coefficient is given byδ (47)If we include damping term in swing equation then equation (43) becomesδδorororδδ . (48)
Where (49)and . . (50)So damped frequency of oscillation,AndTime Constant, T . (51) Example2:Find the maximum steady-state power capability of a system consisting of agenerator equivalent reactance of 0.4pu connected to an infinite bus through aseries reactance of 1.0 p.u. The terminal voltage of the generator is held at1.10 p.u.and the voltage of the infinite bus is 1.0 p.u.Solution:Equivalent circuit of the system is shown in Fig.4.(Fig.4 Equivalent circuit of example2) . (i)δ . (ii)I Using equation (i) and (ii)δδθθδθθθθ . (iii)
Maximum steady-state power apab l tyreal part of equation is zero. Thusrea hedhe δ9 ᵒ, i.e.,θ99p9pLESSON SUMMARY-4:1. Transient Stability-Equal area criterion2. Applications of sudden change in power input3. ExamplesTransient Stability-Equal Area Criterion:The transient stability studies involve the determination of whether or notsynchronism is maintained after the machine has been subjected to severedisturbance. This may be sudden application of load, loss of generation, loss oflarge load, or a fault on the system.A method known as the equal area criterion can be used for a quick predictionof stability. This method is based on the graphical interpretation of the energystored in the rotating mass as an aid to determine if the machine maintains itsstability after a disturbance. This method is only applicable to a one-machinesystem connected to an infinite bus or a two-machine system. Because it providesphysical insight to the dynamic behavior of the machine.Now consider the swing equation (18),oror . (52)
As shown in Fig.5, in an unstable system, δ increases indefinitely with time andmachine looses synchronism. In a stable system, δ undergoes oscillations, whicheventually die out due to damping. From Fig.4, it is clear that, for a system to bestable, it must be that 0 at some instant. This criterion ( 0) can simply beobtained from equation (52).(Fig. 5 A plot of δ (t))Multiplying equation (52) by, we have . (53)This upon integration with respect to time givesdδ . . (54)Where accelerating power and δ is the initial power anglebefore the rotor begins to swing because of a disturbance. The stability ( 0)criterion implies thatdδ . (55)For stability, the area under the graph of accelerating power versus δ must bezero for some value of δ; i.e., the positive (accelerating) area under the graph mustbe equal to the negative (decelerating) area. This criterion is therefore know as theequal area criterion for stability and is shown in Fig. 6.
(Fig.6 Power angle characteristic)Application to sudden change in power input:In Fig. 6 point ‘a’ corresponding to the δ is the initial steady-state operatingpoint. At this point, the input power to the machine,, whereis thedeveloped power. When a sudden increase in shaft input power occurs to , theaccelerating power , becomes positive and the rotor moves toward point ‘b’We have assumed that the machine is connected to a large power system so that Vt does not change and also xd does not change and that a constant field currentmaintains Eg . Consequently, the rotor accelerates and power angle begins toincrease. At pointand δ δ1. Butis still positive and δ overshoots ‘b’,the final steady-state operating point. Now is negative and δ ultimately reaches amaximum value δ2 or point ‘c’ and swing back towards point ‘b’. Therefore therotor settles back to point ‘b’, which is ultimate steady-state operating point.In accordance with equation (55) for stability, equal area criterion requiresArea A1 Area A2orδ dδδdδ . (56)or . (57)ButWhich when substituted in equation (57), we get . . (58)
On simplification equation (58) becomes . . (59)Example 3:A synchronous generator, capable of developing 500MW power per phase,operates at a power angle of 8ᵒ. By how much can the input shaft power beincreased suddenly without loss of stability? Assume that Pmax will remainconstant.Solution:Initially, 8ᵒδ 69.6MW(Fig. 7 Power angle characteristics)Let δm be the power angle to which the rotor can swing before losingsynchronism. If this angle is exceeded, Pi will again become greater than Pe and therotor will once again be accelerated and synchronism will be lost as shown in Fig.7. Therefore, the equal area criterion requires that equation (57) be satisfied withδm replacing δ2.From Fig. 7 δm π-δ1. Therefore equation (59) becomes (i)Substituting 0.139radian in equation (i) gives99 (ii)Solving equation (ii) we get, δ1 50ᵒNowδInitial power developed by machine was 69.6MW. Hence without loss ofstability, the system can accommodate a sudden increase of
9MW per phase 3x313.42 940.3 MW (3-φ) of input shaft power.LESSON SUMMARY-5:1. Critical clearing angle and critical clearing time2. Application of equal area criteriona) Sudden loss of one parallel lineCritical Clearing Angle and Critical Clearing Time:If a fault occurs in a system, δ begins to increase under the influence ofpositive accelerating power, and the system will become unstable if δ becomesvery large. There is a critical angle within which the fault must be cleared ifthe system is to remain stable and the equal area criterion is to be satisfied.This angle is known as the critical clearing angle.(Fig. 8 Single machine infinite bus system)Consider a system as shown in Fig. 8 operating with mechanical inputatsteady angle.as shown by point ‘a’ on the power angle diagram asshown in Fig. 9. Now if three phase short circuit occur at point F of the outgoingradial line , the terminal voltage goes to zero and hence electrical power output ofthe generator instantly reduces to zero i.e.,and the state point drops to ‘b’.The acceleration area A1 starts to increase while the state point moves along bc. At time tc corresponding clearing angle δc, the fault is cleared by the opening ofthe line circuit breaker. tc is called clearing time and δc is called clearing angle.After the fault i s cleared, the system again becomes healthy and transmitspower Pe Pmax sinδ, i.e., the state point shifts to‘d’ on the power anglecurve. The rotor now decelerates and the decelerating area A2 begins to increasewhile the state point moves along d-e. For stability, the clearing a ngle, δc, mustbe such that area A1 area A2.
δ characteristics)(Fig. 9Expressing area A1 Area A2 mathematically we have,δδδδdδδδ dδδδδδδδδδδδ . . (60) . (61)δAlsoUsing equation (60) and (61) we get,δδδδδδδδ (62)Where δ clearing angle, δ initial power angle, and δ power angle towhich the rotor advances (or overshoots) beyond δFor a three phase fault with Pe 0, . (63)Integrating equation (63) twice and utilizing the fact thatδtand t 0 yieldsδ . (64)
If t is the clearing time corresponding to a clearing angle δ , then we obtainfrom equation (64),δtSotδ (65)Note that δ can be obtained from equation (62). As the clearing of faulty line isdelayed, A1 increases and so does δ to find A2 A1 till δδ as shown in Fig.10.(Fig. 10 Critical clearing angle)For a clearing angle (clearing time) larger than this value, the system would beunstable. The maximum allowable value of the clearing angle and clearing time forthe system to remain stable are known as critical clearing angle and criticalclearing time respectively.From Fig. 10,. Substituting this in equation (62) we have,δδδδδδδδδδδδδδδδδ . . . (66)Using equation (65) critical clearing angle can be obtained ast . . . (67)
Application of the Equal Area Criterion:(1) Sudden Loss of One of parallel Lines:PiInfiniteBus V 0ᵒ(a)Eg δPiX1Xd V 0ᵒX2Switched Off(b)(Fig. 11 Single machine tied to infinite bus through two parallel lines)Consider a single machine tied to infinite bus through parallel lines as shown inFig. 11(a). The circuit model of the system is given in Fig. 11(b).Let us study the transient stability of the system when one of the lines issuddenly switched off with the system operating at a steady load. Before switchingoff, power angle curve is given byδδImmediately on switching of line 2, power angle curve is given byδIn Fig. 12, whereinasoperating initially with a steady state power transferon curve I.δ. The system isat a torque angle δ
δ0δ1δ2πδ(Fig. 12 Equal area criterion applied to the opening of one of the two lines inparallel)On switching off line2, the electrical operating point shifts to curve II (point b).Accelerating energy corresponding to area A1is put into rotor followed bydecelerating energy for δ δ1. Assuming that an area A2 corresponding todecelerating energy (energy out of rotor) can be found such that A1 A2, thesystem will be stable and will finally operate at c corresponding to a new rotorangle is needed to transfer the same steady power.If the steady load is increased (line Pi is shifted upwards) a limit is finallyreached beyond which decelerating area equal to A1 cannot be found and therefore,the system behaves as an unstable one. For the limiting case, δ1 has a maximumvalue given byδδδLESSON SUMMARY-6:1. Sudden short circuit on one of parallel linesa) Short circuit at one end of lineb) Short circuit at the middle of a line2. ExampleSudden Short Circuit on One of Parallel Lines:(1) Short circuit at one end of line:Let us a temporary three phase bolted fault occurs at the sending end of one ofthe line.
X1 Eg δXdInfiniteBus V 0ͦᵒPiX2(a)X1 Eg δXdInfiniteBus V 0ͦᵒPiX2(b)(Fig.13 Short circuit at one of the line)Before the occurrence of a fault, the power angle curve is given byδδThis is plotted in Fig. 12.Upon occurrence of a three-phase fault at the generator end of line 2 , generatorgets isolated from the power system for purpose of power flow as shown Fig. 13(b). Thus during the period the fault lasts.The rotor therefore accelerates and angles δ increases. Synchronism will belost unless the fault is cleared in time. The circuit breakers at the two ends of thefaulted line open at time tc (corresponding to angle δc), the clearing time,disconnecting the faulted line. The power flow is now restored via the healthy line(through higher line reactance X2 in place of (), with power angle curveδδ
(Fig. 14 Equal area criterion applied to the system)Obviously,. The rotor now sta
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